I recently started reading The Princeton Companion to Mathematics. I am currently stuck in the introduction to hyperbolic geometry and have some doubts about its models.

Isn't the hyperbolic space produced by rotating a hyperbola? That is, isn't hyperbolic geometry carried out on a hyperboloid?

Also, how is the half plane model generated? Is it generated by taking the crossection of the hyperboloid on the upper half of the complex plane or by taking a projection?

Also, can someone explain to me how "hyperbolic distances become larger and larger, relative to Euclidean ones, the closer you get to the real axis". Is the axis referred to here the real axis of the complex plane or some other axis. And why do distances become larger the closer we get to the real axis? The part of the line close to the real axis simply looks like a part of a circle whose distance does not increase abnormally.

Any visuals will be appreciated.


There are at least four "common" models of the hyperbolic plane:

  1. The "upper" sheet of the hyperboloid of two sheets in Minkowski space (a.k.a., the set of future-pointing unit timelike vectors): $$ x_{1}^{2} + x_{2}^{2} - x_{3}^{2} = -1,\quad x_{3} > 0. $$ Hyperbolic lines turn out to be the intersections of planes through the origin with the hyperboloid.

  2. The Klein disk model, viewed as the unit disk $$ x_{1}^{2} + x_{2}^{2} < 1,\qquad x_{3} = 1, $$ identified with the hyperboloid model by radial projection from the origin. Hyperbolic lines are Euclidean chords. (Patrick Ryan's Euclidean and Non-Euclidean Geometry is a good reference for these two models.)

  3. The Poincaré disk model, viewed as the unit disk $$ x_{1}^{2} + x_{2}^{2} < 1,\qquad x_{3} = 0, $$ identified with the hyperboloid model by radial projection from the point $(0, 0, -1)$ (diagram below). Hyperbolic lines are arcs of Euclidean circles meeting the boundary of the disk orthogonally. Unlike the Klein model, the Poincaré model is conformal; hyperbolic angles coincide with Euclidean angles.

  4. The upper half-plane model, also conformal, obtained from the Poincaré disk model via the fractional linear transformation $$ z \mapsto -i \frac{z + i}{z - i} = \frac{-iz + 1}{z - i}. $$ Hyperbolic lines are Euclidean semicircles (meeting the real axis orthogonally).

Projection from the hyperboloid to the Poincare disk

The Riemannian metrics in the Poincaré and upper half-plane models have well-known formulas in Euclidean coordinates $z = x + iy$: $$ ds^{2} = \frac{4(dx^{2} + dy^{2})}{\bigl(1 - (x^{2} + y^{2})\bigr)^{2}},\qquad ds^{2} = \frac{dx^{2} + dy^{2}}{y^{2}}. $$

Particularly, in the upper half-plane model, a Euclidean distance $ds = \sqrt{dx^{2} + dy^{2}}$ corresponds to a hyperbolic distance $ds/y$; as $y \to 0^{+}$, the hyperbolic length of a Euclidean segment of fixed length grows without bound.

Don Hatch has created an extensive gallery of hyperbolic tessellations (in the Poincaré model) that make this "length distortion" vivid. The "tiles" have fixed hyperbolic shape (and size), and their Euclidean representations shrink toward the boundary of the disk.

Another famous example is the Circle Limit series of prints by M. C. Escher. A web search for "Poincare disk" or "Poincare metric" should turn up many more diagrams.


It looks like you are talking about the Poincaré half-plane model of hyperbolic geometry (see https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model )

Be aware this is only a model of hyperbolic geometry.

For doing "real-life" hyperbolic geometry you need a "surface with a constant negative curvature " or pseudo spherical surface see https://en.wikipedia.org/wiki/Pseudosphere

Also the hyperboloid is only a model of hyperbolic geometry (the curvature of an hyperboloid is positive and not constant, so wrong on both counts of the curvature )

Having said all this you seem to investigate the Poincaré half-plane model of hyperbolic geometry and indeed the scale of this model becomes 0 when you get near the x=0 line

also see


for more information

Hopes this helps


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.